Let S = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let x be the number of 9-digit numbers formed using the digits of the set S such that only one digit is repeated and it is repeated exactly twice. Let y be the number of 9-digit numbers formed using the digits of the set S such that only two digits are repeated and each of these is repeated exactly twice. Then,

Calculating x:

In this case, we have to form a 9-digit number where only one digit is repeated, and it is repeated exactly twice.

Selecting the repeated digit: $$^9C_1=9$$ ways

Now, one digit is used twice, we need 9 - 2 = 7 more distinct digits from the remaining 8 digits in S. This can be selected in $$^8C_7=8$$ ways.

Now, all these 9 digits can be arranged in $$\dfrac{9!}{2!}$$ ways. 

Total possible arrangements: $$9\times8\times\dfrac{9!}{2!}=36\times9!$$ which is equal to x. 

Calculating x:

In this case, we have to form a 9-digit number where two digits are repeated, and each is repeated exactly twice.

Selecting the repeated digits: $$^9C_2=36$$ ways

Now, two digits is used twice, we need 9 - 4 = 5 more distinct digits from the remaining 7 digits in S. This can be selected in $$^7C_5=21$$ ways.

Now, all these 9 digits can be arranged in $$\dfrac{9!}{2!2!}$$ ways.

Total possible arrangements: $$36\times21\times\dfrac{9!}{2!2!}=9\times21\times9!=189\times9!$$ which is equal to y.

To find the relationship, we can divide y by x. 

$$\dfrac{y}{x}=\dfrac{189\times9!}{36\times9!}$$

$$\dfrac{y}{x}=\dfrac{21}{4}$$

Or, we can say,  4y = 21x